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प्रश्न
In the given circle with centre O, PA and PB are tangents and ∠OAB = 28°, then ∠APB is:
विकल्प
90°
56°
62°
90° + 28°
MCQ
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उत्तर
56°
Explanation:
We know that,
Tangent at any point of a circle and the radius through this point are perpendicular to each other.
∴ ∠OAP = 90°
From figure,
⇒ ∠PAB = ∠OAP − ∠OAB
= 90° − 28°
= 62°
PA and PB are tangents drawn to the circle from the external point P.
∴ PA = PB
We know that,
Angles opposite to equal sides of a triangle are equal.
∴ ∠PAB = ∠PBA = 62°
By angle sum property of triangle PAB,
⇒ ∠PAB + ∠PBA + ∠APB = 180°
⇒ 62° + 62° + ∠APB = 180°
⇒ ∠APB = 180° − 124°
= 56°
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